Answer:
71 Ga has a naturally abundance of 36%
Explanation:
Step 1: Given data
Gallium has 2 naturally occurring isotopes: this means the abundance of the 2 isotopes together is 100 %. The atomic weight of Ga is 69.72 amu. This is the average of all the isotopes.
Since the average mass of 69.72 is closer to the mass of 69 Ga, this means 69 Ga will be more present than 71 Ga
Percentage 69 Ga> Percentage 71 Ga
<u>Step 2:</u> Calculate the abundance %
⇒Percentage of 71 Ga = X %
⇒Percentage of 69 Ga = 100 % - X %
The mass balance equation will be:
100*69.72 = x * 71 + (100 - x)*69
6972 = 71x + 6900 -69x
72 = 2x
x = 36 %
71 Ga has a naturally abundance of 36%
69 Ga has a naturally abundance of 64%
Answer:
If the electronegativity difference between bonded atoms are too much high ionic bonds are formed if the electronegativity diference is 0.4 or less than 0.4 non polar covalnet bond formed the difference greater than 0.4 polar covalent bond formed.
Explanation:
Ionic bond:
It is the bond which is formed by the transfer of electron from one atom to the atom of another element.
Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.
For example:
Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion.
Covalent bond:
It is formed by the sharing of electron pair between bonded atoms.
The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.
For example:
In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.
Answer:
![[H^+]=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5x10%5E%7B-13%7DM)
![[OH^-]=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.02M)
Explanation:
Hello there!
In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:
Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:
![\frac{1}{3} =\frac{x}{[Mg(OH)_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%3D%5Cfrac%7Bx%7D%7B%5BMg%28OH%29_2%5D%7D)
Thus, x for this problem is:
![x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5BMg%28OH%29_2%5D%7D%7B3%7D%3D%5Cfrac%7B0.03M%7D%7B3%7D%5C%5C%5C%5Cx%3D%20%200.01M)
Now, according to an ICE table, we have that:
![[OH^-]=2x=2*0.01M=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2x%3D2%2A0.01M%3D0.02M)
Therefore, we can calculate the H^+, pH and pOH now:
![[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B1x10%5E%7B-14%7D%7D%7B0.02%7D%3D5x10%5E%7B-13%7DM)
Best regards!
what?? please reword this
